# Hydraulic force question

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In other words, if you stacked your two one ton car jacks nose to nose,=20 rather than nose to base, you'd be able to lift 2 tons??

In your ASCII diagram, no matter where in the stack you insert a force=20 sensor, it will register the same value. If this were not the case=20 there'd have to be some additional reaction to ground between the rigid=20 surfaces.

Ned Simmons

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[stacked jacks]

They are in series. Sorry, but if you don't know the difference between serial and parallel, further discussion seems quite useless.

Um, ah! So if you stack two jacks and pump one with 1ton you will have 1 ton pressure between the two jacks. Then when you pump the second jack to 1 ton (disregarding it has already a "pressure" of one ton) the force will not increase, but the pressure between will be 2 tons? Get real!

This is my last attempt to explain: First, I ask you to read back a bit and try to understand my posting with the "two opposed spindles C-clamp" example. Thanks.

Now, you have to accept a fact of physics. "actio = reactio". Or, in plain english, "every force has a counter-force (or reacting force)". More verbose: Every force has a reacting force of the same value, but opposed direction and colinear (in line) with that force. Or, the vectorial summ of the forces must be zero. Accept that? OK

Low lets look at the single piston bracket:

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"Bruce L. Bergman" wrote

I can't believe this thread is still going. In your sketch, the vertical _force_ on the rotor is ZERO. That's what you _want_, unless your rotor is on a mighty strong bearing. That's what you _get_, whether you use two "rigid surfaces" and two jacks, or whether you use two "floating surfaces" rigidly connected to each other, and replace one jack with a spacer.

The _horizontal_ force on your rotor is your braking force. It is your friction coefficient times _two_ tons, because there are _two_ sliding interfaces. But there would be two interfaces even if you replace one jack with a spacer and let the caliper float.

Replace one of your "pads" with a (frictionless) _bearing_, and you cut your braking force in half. But the net vertical force _on_ the rotor is still zero, and the net vertical force _through_ the rotor is still one ton.

-- TP

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Did I miss something here or did you mean to respond to different person?

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